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Here are a few recommended readings before getting started with this lesson.
Consider the following statement.
In the applet, rigid motions can be applied only on △ABC.
Reflectbutton.
In the previous exploration, it was seen that a pair of triangles can have corresponding congruent angles but not be congruent triangles. Therefore, relying only on the relationship of only angles is not a valid criterion.
Angle-Angle-Angle is not a valid criterion for proving triangle congruence.
The previous exploration suggests that two triangles are congruent whenever they have two pairs of corresponding congruent sides and the corresponding included angles are congruent. In fact, this conclusion is formalized in the Side-Angle-Side Congruence Theorem
If two sides and the included angle of a triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent.
Based on the diagram above, the theorem can be written as follows.
This proof will be developed based on the given diagram, but it is valid for any pair of triangles.
The primary purpose of the proof is finding a rigid motion or sequence of rigid motions that maps one triangle onto the other. This can be done in several ways. One of the ways will be shown here.In the following diagram, triangles ADE and BCE are congruent, and ∠ADC is congruent to ∠BCD.
How many more pairs of congruent triangles are there in the diagram? Name each congruent triangle pair.
Remember, if two triangles are congruent, then their corresponding sides and angles are congruent.
Start by highlighting the given pair of congruent triangles, △ADE and △BCE.
Since these triangles are congruent, their corresponding parts are congruent. This implies that AD is congruent to BC.
△ADC≅△BCD
△ABD≅△BAC
The last two triangles to consider are triangles ABE and DEC. Unlike the first two pairs, these dimensions seem to be quite different. Therefore, it can be concluded that they are not congruent.
Consequently, in the initial diagram, there are two more pairs of congruent triangles in addition to the given one.
Use segment AB and the rays AX and BY to construct two different triangles, one at a time, in such a way that the following conditions are met.
The following statement could be seen in the previous applet. When two triangles have two pairs of corresponding congruent angles, and the included corresponding sides are congruent, the triangles are then congruent. That leads to the second criteria for triangle congruence.
If two angles and the included side of a triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.
Based on the diagram above, the theorem can be written as follows.
This proof will be developed based on the given diagram, but it is valid for any pair of triangles.
The goal of the proof is to find a rigid motion or sequence of rigid motions that maps one triangle onto the other. This can be done in several ways. One of the ways will be shown here.Consider the following diagram.
What is the value of x+y+z?Take note that QS is a common side for two triangles. Use the fact that if two triangles are congruent, their corresponding sides and angles are congruent.
x=1.5
LHS−1.5=RHS−1.5
LHS/2=RHS/2
Use a calculator
Rearrange equation
As seen in the previous exploration, the Side-Side-Side is a valid criterion for checking triangle congruence.
If the three sides of a triangle are congruent to the three sides of another triangle, then the triangles are congruent.
Based on the diagram above, the theorem can be written as follows.
This proof will be developed based on the given diagram, but it is valid for any pair of triangles.
The primary purpose of this proof is finding a rigid motion or sequence of rigid motions that maps one triangle onto the other. This can be done in several ways. One of them will be shown here.The points C and F′′ are on opposite sides of AB. Now, consider CF′. Let G denote the point of intersection between AB and CF′′.
It can be noted that AC=AF′′ and BC=BF′′. By the Converse Perpendicular Bisector Theorem, AB is a perpendicular bisector of CF′′. Points along the perpendicular bisector are equidistant from the endpoints of the segment, so CG=GF′′.
Finally, F′′ can be mapped onto C by a reflection across AB by reflecting △ABF′′ across AB. Because reflections preserve angles, AF′′ and BF′′ are mapped onto AC and BC, respectively.Given three random segments, it is not always possible to construct a triangle. But, when possible, this triangle will be unique. This fact implies that the angle measures of that triangle are also unique.
Therefore, any other triangle with the same side lengths will also have the same angle measures. Consequently, the two triangles are congruent. Notice that, in contrast, having the same angle measures does not force the side lengths to be unique.In the following diagram, R1 is a rectangle, S1 and S2 are squares, T1, T2, and △MKL are isosceles triangles, DK is congruent to CL, and GM is congruent to JM.
If m∠JCK=3z+8, what is the value of x?Using the Segment Addition Postulate and the Side-Side-Side (SSS) Congruence Theorem, prove that △DGL is congruent to △CJK. Then, find the measure of ∠JCK. Use the fact that m∠JCK+m∠DCB+m∠JCH+x=360∘.
If △DGL and △CJK can be proven to be congruent, that would provide the needed information to find the value of z. Therefore, focus on those two triangles.
DL≅CK
GL≅JK
Since R1 is a rectangle, S1 and S2 are squares, and T1 and T2 are isosceles triangles, the following consequences can be drawn.
Given | Consequence |
---|---|
R1 is a rectangle | DA≅CB |
S1 is a square | DG≅DE |
S2 is a square | CH≅CJ |
T1 is an isosceles triangle | DE≅DA |
T2 is an isosceles triangle | CB≅CH |
Next, organize the information in the right-hand column in a flow chart and use the Transitive Property of Congruence to prove that DG≅CJ.
Notice that the ASA criterion requires the congruent sides to be included between the two pairs of corresponding congruent angles. Using the following applet, investigate what happens when the congruent sides are not the included sides.
Use segment AB and the rays AX and BY to construct two different triangles, one at a time, in such a way that these conditions are met: